Area Related To Circles

We are given that

Area of circle = Area of first circle + Area of second circle

∴ πR² = πR₁² + πR₂²

⇒ R² = R₁² + R₂²

∴ option B is correct.

We are given that

Circumference of circle with radius R = Circumference of first circle with radius R₁ + Circumference of second circle with radius R₂

∴ 2πR = 2πR₁+ 2πR₂

⇒ R = R₁+ R₂

∴ option A is correct.

We are given that

Circumference of a circle = Perimeter of square

Let r be the radius of the circle and a be the side of square.

∴ from the given condition, we have 2π r = 4a

(22/7)r = 2a

⇒ 11r = 7a

⇒ a = (11/7)a

⇒ r = (7/11)a …………..(i)

Now, area of circle = A₁ = πr² and area of square = A₂ = a²

From equation (i ), we have

A₁ = π × (7/11)²

= (22/7) × (49/121)a²

= (14/11)a² and A₂ = a²

∴ A₁ = (14/11) A₂

⇒ A₁ > A₂

Hence, Area of the circle > Area of the square.

A largest triangle that can be inscribed in a semi-circle of radius r units is the triangle having its base as the diameter of the semi-circle and the two other sides are taken by considering a point C on the circumference of the semi-circle and joining it by the end points of diameter A and B.

∴ ∠ C = 90° (by the properties of circle)

So, ΔABC is right angled triangle with base as diameter AB of the circle and height be CD.

Height of the triangle = r

∴ Area of largest ΔABC = (1/2)× Base × Height = (1/2)× AB × CD

= (1/2)× 2r × r = r² sq. units

Let r be the radius of the circle and a be the side of the square.

We are given that Perimeter of a circle = Perimeter of a square

⇒ 2πr = 4a

⇒ a = πr/2

Area of the circle = r² and Area of the square = a²

Now, Ratio of their areas = (Area of circle)/(Area of square)

=

= 4/π = [4/(22/7)] = 14/11

Let D₁ be the diameter of the first circular park = 16m

∴ Radius R₁ of first circular park = 8m

Let D₂ be the diameter of the second circular park = 12m

∴ Radius R₂ of second circular park = 6m

Area of first circular park = πr² = π(8)² = 64 π m²

Area of second circular park = πr² = π(6)² = 36 π m²

Now, we are given that,

Area of single circular park = Area of first circular park + Area of second circular park

∴ πR² = 64 π + 36π = 100π

(where R is the radius of the single circular park)

πR² = 100π ⇒ R² = 100 ⇒ R = 10

∴ Radius of the single circular park will be 10m.

Let a be the side of square = 6 cm

∴ Diameter of a circle = Side of square = 6 cm

∴ Radius of the circle = Diameter/2 = 3 cm

∴ Area of the circle = πr² = π(3)² = 9π cm²

Let r be the radius of circle = OC = 8 cm.

∴ Diameter of the circle = AC = 2 × OC = 2 × 8 = 16 cm

Let a be the side of the square.

Now, according to the given condition,

Diagonal of square = Diameter of the circle.

Now in right angled triangle ACB,

(AC)² = (AB)² + (BC)²

(By Pythagoras theorem)

(16)² = a² + a²

⇒ 256 = 2a²

⇒ a² = 128

∴ Area of the square = a² = 128 cm²

Diameter of first circle = d₁ = 36 cm

Diameter of second circle = d₂ = 20 cm

∴ Circumference of first circle = πd₁ = 36π cm

Circumference of second circle = πd₂ = 20π cm

Now, we are given that,

Circumference of circle = Circumference of first circle + Circumference of second circle

πD = πd₁ + πd₂

⇒ πD = 36π + 20π

⇒ πD = 56π

⇒ D = 56

⇒ Radius = 56/2 = 28 cm

Area of first circle = πr² = π(24)² = 576π m²

Area of second circle = πr² = π(7)² = 49π m²

Now, we are given that,

Area of the circle = Area of first circle + Area of second circle

∴ πR² = 576π +49π

(where, R is the radius of the new circle)

⇒ πR² = 625π

⇒ R² = 625

⇒ R = 25

∴ Radius of the circle = 25cm

Thus, diameter of the circle = 2R = 50 cm.

False

Let a be the side of square.

We are given that the circle is inscribed in the square.

∴ Diameter of circle = Side of square = a

∴ Radius of the circle = a/2

Area of the circle = πr² = π(a/2)² = (πa²)/4 cm²

Hence, area of the circle is (πa²)/4 cm²

True

Let r be the radius of circle = a cm

∴ Diameter of the circle = d = 2 × Radius = 2a cm

As the circle is inscribed in the square, therefore,

Side of a square = Diameter of circle = 2a cm

Hence, Perimeter of a square = 4 × (side) = 4 × 2a = 8a cm

False

Diameter of the circle = d

Therefore,

Diagonal of inner square EFGH = Side of the outer square ABCD = Diameter of circle = d

Let side of inner square EFGH be a

Now in right angled triangle EFG,

(EG)² = (EF)² + (FG)²

By Pythagoras theorem)

⇒ d² = a² +a²

⇒ d² = 2a²

⇒ a² = d²/2

∴ Area of inner circle = a² = d²/2

Also, Area of outer square = d²

∴ the area of the outer circle is only two times the area of the inner circle.

Thus, area of outer square is not equal to four times the area of the inner square.

False

It is not true because in case of major segment, area is always greater than the area of its corresponding sector. It is true only in the case of minor segment.

False

The distance travelled by a circular wheel of radius r in one revolution is equal to the circumference of the circle.

Also, circumference of the circle = 2πd; where d is the diameter of the circle.

True

The distance travelled by a circular wheel of radius r m in one revolution is equal to the circumference of the circle = 2πr

∴ No. of revolutions completed in 2πr m distance = 1

No. of revolutions completed in 1 m distance = (1/2πr)

No. of revolutions completed in s m distance = (1/2πr) × s

False

Let r be the radius of the circle.

Area of the circle = πr²

Circumference of the circle = 2πr

Both are equal only when r = 2 and numerical value of circumference is greater than numerical value of area of circle when 0 < r < 2 and if r > 2, then numerical value of area of circle is greater than the numerical value of the circumference of the circle.

True.

Let P and Q be two circles with radius r and 2r respectively. Let C₁ and C₂ be the centers of the circles P and Q respectively.

Let AB be the arc length of P and CD be the arc length of Q

Let θ₁ and θ₂ be the angle subtended by the arc AB and CD respectively on the center.

Given that AB = CD = l (say)

Now, arc length

CD = = l

∴ from the above two equations, we get

⇒ θ₁ = 2θ₂

∴ Angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.

False

The statement is false because the area of a sector of a circle = (1/2)r²θ, where r is the radius and θ the angle in radians subtended by the arc at the center of the circle. It does not depend on the arc length.

False

Area of first sector = (1/2)(r₁)²θ₁ ,

where r₁ is the radius,

θ₁ is the angle in radians subtended by the arc at the center of the circle.

Area of second sector = (1/2)(r₂)²θ₂,

where r₂ is the radius,

θ₂ is the angle in radians subtended by the arc at the center of the circle.

Given that: (1/2)(r₁)²θ₁ = (1/2)(r₂)²θ₂

⇒ (r₁)²θ₁ = (r₂)²θ₂

It depends on both radius and angle subtended at the center. But arc length only depends on radius of the circle. Therefore, it is not necessary that the corresponding arc lengths are equal. It is possible only if corresponding angles are equal (because then, the corresponding radii will be equal and hence the arc lengths will be equal).

False

The largest circle that can be drawn inside a rectangle is possible when rectangle becomes a square.

∴ Diameter of the circle = Breadth of the rectangle = b

∴ Radius of the circle = b/2

Hence area of the circle = πr² = π(b/2)²

True

We are given that

Circumference of circle with radius R₁ = Circumference of circle with radius R₂

⇒ 2πR₁= 2πR₂

⇒ R₁= R₂

⇒ π(R₁)² = π(R₂)²

and hence the areas are also equal.

True

We are given that

Area of circle with radius R₁ = Area of circle with radius R₂

⇒ π(R₁)² = π(R₂)²

⇒ R₁= R₂

⇒ 2πR₁= 2πR₂, and hence the circumferences are also equal.

False

When the square is inscribed in the circle, the diameter of a circle is equal to the diagonal of a square but not the side of the square.

Let side of square = a

∴ p²= a² +a²

⇒ p²= 2a²

⇒ a² = p²/2

Hence area of square = a² = p²/2

Radius of first circle = r₁ = 15 cm

Radius of second circle = r₂ = 18 cm

∴ Circumference of first circle = 2πr₁ = 30π cm

Circumference of second circle = 2πr₂ = 36π cm

Let R be the radius of the circle.

Now, we are given that,

Circumference of circle = Circumference of first circle + Circumference of second circle

2πR= 2πr₁+ 2πr₂

⇒ 2πR = 30π + 36π

⇒ 66π ⇒ R = 33

⇒ Radius = 33 cm

Hence, required radius of a circle is 33 cm.

Let a be the side of square.

∴ Diameter of a circle = Diagonal of the square = 8 cm

Now in right angled triangle ABC,

(AC)² = (AB)² + (BC)²

(By Pythagoras theorem)

∴ (8)²= a² +a²

⇒ 64= 2a²

⇒ a²= 32

Hence area of square = a²= 32 cm²

∴ Radius of the circle = Diameter/2 = 4 cm

∴ Area of the circle = πr² = π(4)² = 16 cm²

So, the area of the shaded region = Area of circle – Area of square

∴ the area of the shaded region = 16π – 32

= 16 × (22/7) – 32

= 128/7

= 18.286 cm²

Area of a sector of a circle = (1/2)r²θ,

where r is the radius and,

θ the angle in radians subtended by the arc at the center of the circle

Here, Radius of circle = 28 cm

Angle subtended at the center = 45°

Angle subtended at the center (in radians) = θ 45π/180 = π/4

∴ Area of a sector of a circle =

=

= 308 cm²

Hence, the required area of a sector of a circle is 308 cm^2.

Radius of wheel = r = 35 cm

1 revolution of the wheel = Circumference of the wheel

= 2πr

= 2 × (22/7) × 35

= 220 cm

But Speed of the wheel = 66 km/hr

= cm/min

= 110000 cm/min

∴ Number of revolutions in 1 min = 110000/220 = 500

Hence, required number of revolutions per minute is 500.

Let ABCD be a rectangular field.

Length of field = 20 m

Breadth of the field = 16 m

Suppose a cow is tied at a point A.

Let length of rope be AE = 14 m = l (say).

Angle subtended at the center of the sector = 90°

Angle subtended at the center (in radians) θ = 90π/180 = π/2

∴ Area of a sector of a circle

= 154 m²

Hence, the required area of a sector of a circle is 154 m^2.

Length and breadth of the rectangular portion AFDC of the flower bed are 38 cm and 10 cm respectively.

Area of the flower bed = Area of the rectangular portion + Area of the two semi-circles.

∴ Area of rectangle AFDC = Length × Breadth

= 38 × 10 = 380 cm²

Both ends of flower bed are semi-circle in shape.

∴ Diameter of the semi-circle = Breadth of the rectangle AFDC = 10 cm

∴ Radius of the semi circle = 10/2 = 5 cm

Area of the semi-circle = πr²/2 = 25π/2 cm²

Since there are two semi-circles in the flower bed,

∴ Area of two semi-circles = 2 × (πr²/2) = 25π cm²

Total area of flower bed = (380 + 25π) cm²

Given, AC = 6cm and BC = 8 cm

A triangle in a semi-circle with hypotenuse as diameter is right angled triangle.

∴ In right angled triangle ACB,

(AB)² = (AC)² + (CB)²

(By Pythagoras theorem)

(AB)² = (6)² + (8)²

⇒(AB)² = 36 + 64

⇒(AB)² = 100 ⇒(AB)= 10

∴ Diameter of the circle = 10 cm

Thus, Radius of the circle = 5 cm

Area of circle = πr²

= π(5)²

= 25π cm²

= 25 × 3.14 cm²

= 78.5 cm²

Also, Area of the right angled triangle = (1/2) × Base × Height

= (1/2) × AC × CB

= (1/2) × 6 × 8 = 24 cm²

Now, Area of the shaded region = Area of the circle – Area of the triangle

= (78.5-24)cm²

= 54.5cm²

We can clearly see that the figure comprises of a rectangle and a semi-circle.

Name the rectangle ABCD. Name the point where the

rectangle and the semi-circle meets as E.

Take any point F on the semi-circle, then we have

the semi-circle as EFD.

Now, area of the figure = Area of the semi-circle + Area of the rectangle.

Here, from the figure, Radius of the semi-circle = r = 6 - 4 = 2 m

∴ Area of the semi-circle = πr²/2 = 4π/2 = 2π

Also, area of the rectangle = Length × Breadth

= AB × BC

= 4 × 8 = 32 m²

∴Area of shaded region = Area of rectangle ABCD + Area of semi-circle DEF

= (32 +2π) m²

Label the figure as above.

Area of the shaded region = Area of the rectangle ABCD – (Area of the rect. EFGH + (Area of the semi-circle EFJ +Area of the semi-circle GHI)

Length and breadth of outer rect. ABCD are 26 m and 12 m respectively.

∴ Area of the rect. ABCD = Length × Breadth
= AB × BC
= 26 × 12
= 312 m²

From the figure, length and breadth of inner rect. EFGH are (26-5-5) m and (12-4-4) m, i.e. 16 m and 4 m respectively.

∴ Area of the rect. EFGH = Length × Breadth

= EF × FG

= 16 × 4
= 64 m²

Breadth of the inner rectangle = Diameter of the semi-circle EJF = d = 4m

∴ Radius of semi-circle EJF = r = 2 m

Area of the semi-circle EFJ = Area of the semi-circle GHI
= πr²/2
= 4π/2
= 2π m²

∴ Area of shaded region = 312 – (64 + 2π + 2π)m²
= ( 248 – 4π ) m²

Let r be the radius of the circle = 14 cm

Angle subtended at the center of the sector = θ = 60°

In triangle AOB, ∠AOB = 60°, ∠OAB = ∠OBA = θ

Since, sum of all interior angles of a triangle is 180°

∴ θ + θ + 60 = 180

⇒ 2 θ = 120

⇒ θ = 60

∴ Each angle is of 60° and hence the triangle AOB is an equilateral triangle.

Now, Area of the minor segment = Area of the sector AOBC – Area of triangle AOB

Angle subtended at the center of the sector = 60°

Angle subtended at the center (in radians) = θ = 60π/100 = π/3

∴ Area of a sector of a circle = r²θ/2

=

= 308/3 cm²

Area of the equilateral triangle =

∴ Area of minor segment = = 17.796 cm²

Since P, Q, R and S divides AB, BC, CD and DA in half.

∴ AP = PB = BQ = QC = CR = RD = DS = SA = 6 cm.

Given, side of a square BC = 12 cm

Area of the square = 12 × 12 = 144 cm²

Area of the shaded region = Area of the square - (Area of the four quadrants)

Area of one quadrant = (π/4)×(Radius)² = (3.14/4)× 36 = 113.04/4 cm²

Area of four quadrants = 113.04 cm²

Area of the shaded region = 144-113.04 = 30.96 cm²

Since D, E, F bisects BC, CA, AB respectively.

∴ AE = EC = CD = DB = BF = FA = 5 cm

Now area of the shaded region = (Area of the three sectors)

Since the triangle is an equilateral triangle, therefore each angle is of 60°

∴ Angle subtended at the center of each sector = 60°

Angle subtended at the center (in radians) = θ = 62π/180 = π/3

Radius of each sector = 5 cm

∴ Area of a sector of a circle

∴ Area of three sectors of a circle

= 78.5/2 cm²

= 39.25 cm²

∴ Area of shaded region = 39.25 cm²

Let r be the radius of each sector = 14 cm

Area of the shaded region = Area of the three sectors

Let angles subtended at P, Q, R be x°, y°, z° respectively.

Angle subtended at P, Q, R (in radians, (θ)) be respectively.

∴ Area of a sector with central angle at P

∴ Area of a sector with central angle at Q

∴ Area of a sector with central angle at R =

∴ Area of three sectors =

Since, sum of all interior angles in any triangle is 180°

∴ x + y + z = 180°

Thus, Area of three sectors = = 308 cm²

Hence, the required area of the shaded region is 308 cm².

Let r be the radius of the park = 105 m

Given that the circular park is surrounded by a road of width 21 m.

So, Radius of the outer circle = R = (105+21) m = 126 m

Area of the road = Area of the outer circle – Area of the circular park

= πr² – πR²

=

=

= 15246 m²

Hence, the required area of the road is 15246 m².

Let r be the radius of each sector = 21 cm

Area of the shaded region = Area of the four sectors

Let angles subtended at A, B, C and D be x°, y°, z° and w° respectively.

Angle subtended at A, B, C, D (in radians, (θ)) be respectively.

∴ Area of a sector with central angle at A =

∴ Area of a sector with central angle at B

∴ Area of a sector with central angle at C

∴ Area of a sector with central angle at D

∴ Area of four sectors =

Since, sum of all interior angles in any quadrilateral is 360°

∴ x + y + z +w = 360°

Thus, Area of four sectors =

= 441π cm²

= 1386 cm²

Hence, required area of the shaded region is 1386 cm².

Area of the circular playground = 22176 m² (Given)

Let r be the radius of the circle.

∴ πr² = 22176

⇒ (22/7)r² = 22176

⇒ r² = 22176 × (22/7)

⇒ r² = 7056

⇒ r = 84

∴ Radius of the circular playground = 84 m

Now, circumference of the circle = 2πr

= 2×(22/7)×84

= 528 m

Cost of fencing 1 meter of ground = Rs 50

∴ Cost of fencing the total ground = Rs 528 × 50 = Rs 26,400

Diameter of front wheels = d₁ = 80 cm

Diameter of rear wheels = d₂ = 2 m = 200 cm

Let r₁ be the radius of the front wheels = 80/2 = 40 cm

Let r₂ be the radius of the rear wheels = 200/2 = 100 cm

Now, Circumference of the front wheels = 2πr

= 2 ×(22/7) × 40

= 1760/7 cm

Circumference of the rear wheels = 2πr = 2 × (22/7) × 100 = 4400/7 cm

No. of revolutions made by the front wheel = 1400

∴ Distance covered by the front wheel = 1400 × (1760/7) = 352000 cm

Number of revolutions made by rear wheel in covering a distance in which the front wheel makes 1400 revolutions =

=

= 560 revolutions.

Sides of the triangle are 15 m, 16 m, and 17 m.

Now, perimeter of the triangle = (15+16+17) m = 48 m

∴ Semi-perimeter of the triangle = s = 48/2 = 24 m

By Heron’s formula, Area of the triangle =

(where a, b, c are the sides of triangle)

=

= 109.982 m²

Let B, C and H be the corners of the triangle on which buffalo, cow and horse are tied respectively with ropes of

7 m each.

So, the area grazed by each animal will be in the form of a sector.

∴ Radius of each sector = r = 7 m

Let x, y, z be the angles at corners B, C, H respectively.

∴ Area of sector with central angle x =

Area of sector with central angle y =

Area of sector with central angle z =

Area of field not grazed by the animals = Area of triangle – (area of the three sectors)

=

=

= 109.892 – 77 = 32.982 cm²

Radius of the circle = r = 12 cm

∴ OA = OB = 12 cm

∠ AOB = 60° (given)

Since triangle OAB is an isosceles triangle, ∴ ∠ OAB = ∠ OBA = θ (say)

Also, Sum of interior angles of a triangle is 180°,

∴ θ + θ + 60° = 180°

⇒2θ = 120° ⇒ θ = 60°

Thus, the triangle AOB is an equilateral triangle.

∴ AB = OA = OB = 12 cm

Area of the triangle AOB = × a² ,

where a is the side of the triangle.

× (12)²

= (√3/16) ×144

=

= 62.354 cm²

Now, Central angle of the sector AOBCA = = 60° = = (π/3) radians

Thus, area of the sector AOBCA =

= = 75.36 cm²

Now, Area of the segment ABCA = Area of the sector AOBCA – Area of the triangle AOB

= (75.36 – 62.354) cm² = 13.006 cm²

Diameter of the circular pond = 17.5 m

Let r be the radius of the park = (17.5/2) m = 8.75 m

Given that the circular pond is surrounded by a path of width 2 m.

So, Radius of the outer circle = R = (8.75+2) m = 10.75 m

Area of the road = Area of the outer circular path – Area of the circular pond

= πr² – πR²

= 3.14 × (10.75)² – 3.14 × (8.75)²

= 3.14 × ((10.75)² – (8.75)²)

= 3.14 × ((10.75 + 8.75) × (10.75 – 8.75))

= 3.14 × 19.5 × 2 = 122.46 m²

Hence, the area of the path is 122.46 m².

Now, Cost of constructing the path per m² = Rs. 25

∴ cost of constructing 122.46m² of the path = Rs. 25 × 122.46 = Rs. 3061.50

AB = 18 cm, DC = 32 cm

Distance between AB and DC = Height = 14 cm

Now, Area of the trapezium = (1/2) × (Sum of parallel sides) × Height

= (1/2) × (18+32) × 14 = 350cm²

As AB ∥ DC, ∴ ∠ A +∠ D = 180°

and ∠ B +∠ C = 180°

Also, radius of each arc = 7 cm

Therefore,

Area of the sector with central angle A = (1/2) × (∠A/180) × π × r²

Area of the sector with central angle D = (1/2) × (∠D/180) × π × r²

Area of the sector with central angle B (1/2) × (∠B/180) × π × r²

Area of the sector with central angle C = (1/2) × (∠C/180) × π × r²

Total area of the sectors =

= 77 + 77 = 154

∴ Area of shaded region = Area of trapezium – (Total area of sectors)

= 350 – 154 = 196 cm²

Hence, the required area of shaded region is 196 cm².

The three circles are drawn in such a way that each of them touches the other two.

So, by joining the centers of the three circles, we get,

AB = BC = CA = 2(Radius) = 7 cm

Therefore, triangle ABC is an equilateral triangle with each side 7 cm.

∴ Area of the triangle× a²

where a is the side of the triangle.

= 21.2176 cm²

Now, Central angle of each sector = = 60° (60π/180)

= π/3 radians

Thus, area of each sector = (1/2) r²θ

= (1/2) × (3.5)² × (π/3)

=

= 6.4167 cm²

Total area of three sectors = 3 × 6.4167 = 19.25 cm²

∴ Area enclosed between three circles = Area of triangle ABC – Area of the three sectors

= 21.2176 – 19.25

= 1.9676 cm²

Hence, the required area enclosed between these circles is 1.967 cm² (approx.).

Radius of the circle = r = 5 cm

Arc length of the sector = l = 3.5 cm

Let the central angle (in radians) be θ.

As, Arc length = Radius × Central angle (in radians)

∴ Central angle (θ) = Arc length / Radius = l / r = 3.5/5 = 0.7 radians

Now, Area of the sector = (1/2) × r²θ = (1/2) × 25 × 0.7 = 8.75 cm²

Hence, required area of the sector of a circle is 8.75 cm².

The four circles are placed in such a way that each piece touches the other two pieces.

So, by joining the centers of the circles by a line segment, we get a square ABDC with sides as,

AB = BD = DC = CA = 2(Radius) = 2(7) cm = 14 cm

Now, Area of the square = (Side)² = (14)² = 196 cm²

Since, ABDC is a square, ∴ each angle has a measure of 90°.

∴ ∠ A = ∠ B = ∠ D = ∠ C = 90° = π/2 radians = θ (say)

Also, Radius of each sector = 7 cm

Thus,

Area of the sector with central angle A = (1/2)r²θ

=

=

= (77/2) cm²

Since the central angles and the radius of each sector are same, therefore area of each sector is 77/2 cm²

∴ Area of the shaded portion = Area of square – Area of the four sectors

= 196 – 154

= 42 cm²

Hence, required area of the portion enclosed between these pieces is 42 cm².

Since, Area of the square = 784 cm²

∴ Side of the square = √Area = √784 = 28 cm

Since the four circular plates are congruent, therefore diameter of each circular plate = 28/2 = 14 cm

∴ Radius of each circular plate = 7 cm

Area of the sheet not covered by plates =

Area of the square – Area of the four circular plates

Since all four circular plates are congruent, therefore area of all four plates will be equal.

∴ Are of one circular plate = πr² = = 154 cm²

So, Area of four plates = 4×154 = 616 cm²

Area of the sheet not covered by plates = 784 – 616 = 168 cm²

Length of the floor = l = 5 m

Breadth of the floor = b = 4 m

∴ Area of the floor = l × b = 5 × 4 = 20 m²

Now, Diameter of each circular tile = 50 cm

∴ Radius of each circular tile = r = 25 cm = 0.25 m

Area of one circular tile = πr²

= 3.14 × (0.25)²

= 0.19625 m²

Area of such 80 tiles = 80 × 0.19625

= 15.7 m²

Area of the floor that remains uncovered with tiles =

Area of the floor – Area of all 80 circular tiles

= 20 – 15.7 = 4.3 m²

Hence, the required area of floor that remains uncovered with tiles is 4.3 m².

Are of the circle = 1256 cm²

Let r be the radius of the circle.

∴ Area = πr²⇒ 1256 = 3.14 × r²⇒ r² = 1256/3.14 = 400

⇒ r = 20 cm

∴ Diameter of the circle = d = 2×20 = 40 cm

Since all the vertices of a rhombus lie on a circle, therefore the diagonals of the rhombus pass through the center of the circle and thus diagonals of the rhombus are equal to the diameter of the circle.

Let d₁ and d₂ be the diagonal of the rhombus.

Since, diagonals of a rhombus are equal, therefore d₁ = d₂ = d = 40 cm

Now, Area of the rhombus = (1/2) × d₁ × d₂

= (1/2) × 40 × 40 = 800 cm²

Hence, the required area of rhombus is 800 cm².

Diameters are in the ratio 1:2:3

So, let the diameters of the concentric circles be 2r, 4r and 6r.

∴ Radius of the circles be r, 2r, 3r respectively.

Now, Area of the outermost circle = π (Radius)² = π (3r)² = 9πr²

Area of the middle circle = π (Radius)² = π (2r)² = 4πr²

Area of the innermost circle = π (Radius)² = π (r)² = πr²

Now, Area of the middle region = Area of middle circle – Area of the innermost circle

= 4πr² - πr² = 3πr²

Now, Area of the outer region = Area of outermost circle – Area of the middle circle

= 9πr² - 4πr² = 5πr²

Required ratio = Area of inner circle: Area of the middle region: Area of the outer region

= πr² : 3πr² : 5πr²

⇒ Required Ratio is 1:3:5

Length of the minute hand = 5 cm = Radius of the clock

Minutes between the time period 6:05 am to 6:40 am = 35 minutes

In 60 minutes, the minute hand completes one revolution, i.e. 360°.

∴ Angle made by minute hand in 1 minute = 360°/60° = 60°

Thus angle made by minute hand in 35 minutes = 60° × 35 = 210°

Angle made by minute hand in 35 minutes (in radians) = θ = (210π/180)

∴ Area swept by minute hand in 35 minutes =

= 275/6

= 45.833 cm²

Hence, the required area swept by the minute land is 45.833 cm²

Let r be the radius of the circle.

Given, Central angle = 200° = (200π/180) radians = θ (say)

Thus, Area of the sector = 770 cm²

⇒ 770 = (1/2)r²θ

= 9 ×49

⇒ r = 21 cm

Thus radius of the sector = 21 cm

Now, Length of the corresponding arc = Radius Central angle (in radians)

=

= 220/3 cm = 73.33 cm

Hence, the required length of the corresponding arc is 73.33 cm.

Radius of one sector = r₁ = 7 cm

Radius of second sector = r₂ = 21 cm

Central angle of one sector = 120°

Central angle of second sector = 40°

Central angle of one sector (in radians) = θ₁ = (120π/180)

Central angle of second sector (in radians) = θ₂ = (40π/180)

Area of first sector =

= =

154/3 = 51.33 cm²

Area of second sector =

=

= 154 cm²

Let the lengths of the corresponding arc be l₁ and l₂.

Now, arc length of first sector = Radius × Central Angle (in radians)

= = 44/3 cm

Now, arc length of second sector = Radius × Central Angle (in radians)

= = 44/3 cm

Hence, we observe that arc lengths of two sectors of two different circles may be equal but their area need not be equal.

Area of square ABCD = side²
= 14²
= 196 cm²
As,

14 = 3 + r + 2r + r + 3

⇒ 14 = 6 + 4r

⇒ 14 – 6 = 4r

⇒ 8 = 4r

⇒ 2 cm = r

Area of internal portion = Area of 4 semi-circles + Area of square JKLM ..... (1)
Area of semi-circle =



Area of 4 semicircles is:
=

Side of square = 2r
= 2(2)
= 4 cm
Area of square = side²= 4²
= 16 cm²
From (1),
Area of the internal portion
Area of shaded region = Area of square ABCD - Area of the internal portion





= 154.85 cm²

Let r be the radius of the circular wheel.

Area = 1.54 m²

∴πr² = 1.54

= 0.49

⇒ r = 0.7 m

Thus, radius of the circular wheel = 0.7 m

Circumference of the wheel = 2πr = 2 × (22/7) × 0.7 = 4.4 m

Distance travelled by wheel in one revolution = Circumference of circular wheel = 4.4 m

Since, distance travelled by a circular wheel = 176 m

Total distance covered by the wheel = No. of revolutions made by wheel × (Distance

covered in one revolution)

∴ No. of revolutions made by wheel =

= 176/4.4

= 40

Hence, the required number of revolutions made by a circular wheel is 40.

Length of the chord = 5 cm (Given)

Let r be the radius of the circle.

Then, OA = OB = r cm

Now, angle subtended at the center of the sector OABO = 90°

Angle subtended at the center of the sector OABO (in radians) = θ = π/2

∴ Triangle AOB is a right-angled triangle.

So, by Pythagoras theorem, (AB)² = (OA)² + (OB)²

⇒ 25 = 2r²

Also, AOB is an isosceles triangle.

Since, line segment OD is perpendicular on AB, therefore it divides Ab into two equal parts. Thus, AD = DB = 5/2 = 2.5 cm

Let AD = h cm

So, in right angled triangle AOD, by Pythagoras theorem,

(AO)² = (AD)² + (OD)²

= 25/4

⇒ h = 5/2 = 2.5 cm

∴ Area of the isosceles triangle AOB (1/2) × Base × Height

= (1/2) × 5 × (5/2) = 25/4 cm²

Now, Area of the minor sector = (1/2)r²θ

= = (25π/8) cm²

Area of the minor segment = Area of the minor sector – Area of the isosceles triangle

=

Area of the major segment = Area of the circle – Area of the minor segment

=

=

∴ Difference of the areas of two segments of a circle =

|Area of major segment – Area of minor segment|

Hence, the required difference of the areas of two segments is

Radius of the circle = r = 21 cm

Area of the circle

Central angle of the sector AOBA = 120°

Central angle of the sector AOBA (in radians) = θ (120π/180) = 2π/3

Now, area of the minor sector AOBA = (1/2)r²θ

= (1/2) × (21)² × (2π/3)

= 462 cm²

Area of the major sector ABOA = Area of the circle – Area of the sector AOBA

= 1386 – 462 = 924 cm²

Now, Difference of the areas of a sector AOBA and its corresponding major sector ABOA

= |Area of major sector ABOA – Area of minor sector AOBA|

= |924-462| = 462 cm²

Hence, the required difference of two sectors is 462 cm².